JEE Physics System Of Particles and Rotational Motion MCQs Set 39

Answer: (c) Pairs 3 and 4 only

Explanation:
1. The angle turned must lie between \( \pi \approx 3.14 \) rad and \( 2\pi \approx 6.28 \) rad.
2. Differences: Pair 1 gives 3 rad, Pair 2 gives 2 rad, Pair 3 gives 4 rad, Pair 4 gives 3.5 rad.
3. Only Pairs 3 and 4 fall in the allowed range.

Teacher's Note:
1. Convert the limits 180 and 360 degrees to radians and compare each angular displacement (final minus initial) with them.

Answer: (a) 1 and 3 only

Explanation:
1. Torque about B: \( N_A d = w(d-x) \), so \( N_A=\dfrac{w(d-x)}{d} \).
2. Torque about A: \( N_B d = wx \), so \( N_B=\dfrac{wx}{d} \).
3. Hence statements 1 and 3 are correct.

Teacher's Note:
1. Take moments about each knife edge in turn; the nearer support carries the larger share.

Answer: (a) 1, 2 and 3 only

Explanation:
1. Acceleration: \( a=\dfrac{(7-4)g}{4+7+3}=\dfrac{3g}{14} \) (pulley effective mass \( M/2=3 \) kg).
2. Tensions: \( T_{4}=4(g+a)=\dfrac{34g}{7}\approx4.86g \), \( T_{7}=7(g-a)=\dfrac{11g}{2}=5.5g \), so statement 1 is true.
3. Pulley KE \( =\tfrac12\cdot\tfrac12(6)R^2\omega^2=1.5v^2 \) and block KE \( =2v^2 \), so statement 2 is true.
4. Hinge force \( =6g+T_4+T_7\approx16.4g<17g \), so statement 3 is true.
5. After the 7 kg block stops, the pulley keeps turning and pulls the 4 kg side taut, so the string does not go slack (statement 4 false).

Teacher's Note:
1. Treat the disc pulley as an effective mass M/2 in the linear equation, then check each statement with the tensions and the hinge force.

Answer: (a) 1, 3 and 4 only

Explanation:
1. With no external torque, angular momentum \( L=I\omega \) stays constant.
2. Changing the shape changes \( I \), so \( \omega=L/I \) changes and \( K=\dfrac{L^2}{2I} \) changes.
3. Hence 1, 3 and 4 change; angular momentum does not.

Teacher's Note:
1. Angular momentum is conserved when no external torque acts; everything else depends on I and can vary.

Answer: (a) 1 and 2 only

Explanation:
1. The tension acts through the hole, so angular momentum is conserved: \( mv_1r_1=mv_2r_2 \Rightarrow v_2=\dfrac{1\times1}{0.5}=2 \) m/s.
2. New tension: \( T=\dfrac{mv_2^2}{r_2}=\dfrac{0.1\times4}{0.5}=0.8 \) N.
3. So statements 1 and 2 are correct.

Teacher's Note:
1. A central force gives no torque about the hole, so use conservation of angular momentum first and then the circular motion equation.

Answer: (b) 1, 3 and 4 only

Explanation:
1. Speed of a point at angle \( \phi \) measured from the contact point (at the centre) is \( v=2u\sin(\phi/2) \), which ranges from 0 to 2u (1 true).
2. CP horizontal means \( \phi=90^\circ \): \( v=2u\sin45^\circ=\sqrt2\,u \), so 2 is false and 4 is true.
3. CP at 30 degrees below horizontal means \( \phi=60^\circ \): \( v=2u\sin30^\circ=u \), so 3 is true.

Teacher's Note:
1. The velocity of any point is the vector sum of u and the rotational part, which gives v = 2u sin(phi/2).

Answer: (b) It may start pure rolling after some time, or it may never start pure rolling.

Explanation:
1. Kinetic friction acts up the incline and increases the angular speed while reducing the linear acceleration.
2. If \( \mu \ge \tfrac13\tan\theta \) (solid cylinder) the contact point eventually stops slipping and pure rolling begins.
3. If \( \mu \) is smaller, the cylinder slides throughout and never rolls purely, so both outcomes are possible.

Teacher's Note:
1. Whether rolling is reached depends on the friction coefficient compared with the incline angle.

Answer: (b) 2, 3 and 4 only

Explanation:
1. Rolling part: \( a_1=\dfrac{2F}{3m} \), \( s=\tfrac12a_1T^2 \), \( v_B=a_1T \).
2. On the smooth part the torque is zero so \( \omega \) is constant while \( a_2=\dfrac Fm=1.5a_1 \): the linear acceleration increases (2 true, 1 false).
3. One rotation takes \( \dfrac{2\pi R}{v_B}=\dfrac{s}{a_1T}=\dfrac T2 \) (3 true).
4. Distance in time T: \( v_BT+\tfrac12a_2T^2=1.75a_1T^2=3.5s>s \) (4 true).

Teacher's Note:
1. Without friction the disc keeps its spin and accelerates faster, since all the force now goes into linear motion.

Answer: (d) 1, 2 and 3

Explanation:
1. The sphere accelerates because of friction from the plank, and its rolling relative to the plank means its centre moves slower than the plank (1 true).
2. Friction is the only horizontal force on the sphere, so the work done on it equals its gain in kinetic energy (2 true).
3. There is no slipping, so no energy is dissipated, and the ground is smooth, so the work done by F becomes total kinetic energy (3 true).

Teacher's Note:
1. With no slipping there is no dissipation, so the work of the external force is all stored as kinetic energy.

Answer: (c) \( \dfrac83 mvR \)

Explanation:
1. The non-viscous liquid does not rotate; only the shell (hollow sphere, \( I=\tfrac23mR^2 \)) spins at \( \omega=v/R \).
2. About a ground point: \( L=(2m)vR+I\omega=2mvR+\tfrac23mR^2\cdot\dfrac vR \).
3. So \( L=\dfrac83mvR \).

Teacher's Note:
1. Add the orbital part of the total mass to the spin of the shell only, since the ideal liquid does not take part in the rotation.

Answer: (a) 1, 3 and 4 only

Explanation:
1. Torque of gravity about O is \( mg\times \) (horizontal distance), which is constant, so \( L \) increases linearly (1 true, 2 false).
2. The distance from O decreases as the particle falls, so \( I=mr^2 \) decreases (3 true).
3. \( \omega=L/I \) therefore increases (4 true).

Teacher's Note:
1. The gravitational torque stays constant because the horizontal lever arm does not change.

Answer: (a) 1, 2 and 3 only

Explanation:
1. About B: \( L_B=I_B\omega=\tfrac32mR^2\omega \), clockwise.
2. About A: the spin part is \( \tfrac12mR^2\omega \) clockwise and the orbital part is \( mR\cdot v=mR^2\omega \) anticlockwise, so \( L_A=\tfrac12mR^2\omega \) anticlockwise.
3. Ratio \( L_B:L_A=3:1 \), and the senses are opposite, so statements 1, 2 and 3 are correct.

Teacher's Note:
1. Use L = L(cm) + r x mv about each point, watching the senses of rotation.

Answer: (a) 1, 2 and 3 only

Explanation:
1. The contact point slides forward, so kinetic friction acts backward (1, 2 true).
2. Backward friction at the bottom gives a clockwise torque about the centre for a ring moving right (3 true).
3. Rolling begins only when \( v=R\omega \), after some time, not immediately (4 false).

Teacher's Note:
1. Friction opposes the sliding of the contact point; it slows the centre and spins the ring up.

Answer: (c) The angular momentum of the ring about any point on the horizontal surface

Explanation:
1. Friction is an external force, so linear momentum changes (a false).
2. Friction gives a torque about the centre, so angular momentum about the centre changes (b false).
3. About a point on the surface, friction and normal force have zero moment (their lines lie on the surface line) and the weight and normal force balance, so angular momentum is conserved (c true).
4. Friction dissipates energy, so mechanical energy falls (d false).

Teacher's Note:
1. Choosing a point on the line of action of friction removes its torque, so angular momentum about such a point is conserved.

Answer: (a) 1, 3 and 4 only

Explanation:
1. Rolling starts when the contact point is at rest, \( v=R\omega \) (1 true, 2 false).
2. \( v=v_0-\mu gt \), \( R\omega=\mu gt \) (angular acceleration \( \mu g/R \) for a ring). Equating: \( t=\dfrac{v_0}{2\mu g} \) (3 true).
3. Then \( v=\dfrac{v_0}2 \) (4 true).

Teacher's Note:
1. Set the linear speed equal to R times the angular speed and solve for the time.

Answer: (b) 1, 3 and 4 only

Explanation:
1. Before rolling: \( s=v_0t-\tfrac12\mu gt^2 \) with \( t=\dfrac{v_0}{2\mu g} \), giving \( s=\dfrac{3v_0^2}{8\mu g} \) (1 true).
2. Final \( v=\dfrac{v_0}{2} \): total KE \( =2\times\tfrac12M\left(\dfrac{v_0}2\right)^2=\dfrac{Mv_0^2}{4} \), so the loss is \( \dfrac12Mv_0^2-\dfrac14Mv_0^2=\dfrac{Mv_0^2}{4} \) (3 true) and the work done by friction is \( -\dfrac{Mv_0^2}{4} \) (2 false).
3. Rotational KE gained \( =\tfrac12MR^2\left(\dfrac{v_0}{2R}\right)^2=\dfrac{Mv_0^2}{8} \) (4 true).

Teacher's Note:
1. Work done by friction equals the change in total kinetic energy, including the rotational part.

Answer: (a) 1, 2 and 3 only

Explanation:
1. \( \dfrac{d\vec L}{dt}=\vec A\times\vec L \) is perpendicular to both \( \vec A \) and \( \vec L \) (1 true).
2. Perpendicular to \( \vec A \) means the component of \( \vec L \) along \( \vec A \) is constant (2 true).
3. Perpendicular to \( \vec L \) means \( |\vec L| \) is constant (3 true), but the direction of \( \vec L \) keeps changing, so 4 is false.

Teacher's Note:
1. A derivative that is always perpendicular to a vector changes only its direction, not its magnitude.

Answer: (a) 2, 3 and 4 only

Explanation:
1. About O the vector \( \vec L \) has constant magnitude and constant vertical component but its horizontal component rotates, so it is not constant as a vector (1 false, 3 and 4 true).
2. About C the angular momentum is \( mvr \) directed along the vertical axis, which is constant (2 true).

Teacher's Note:
1. A rotating vector can have constant magnitude and constant vertical component while its direction still changes.

Answer: (a) 1 and 3 only

Explanation:
1. Rolling friction slows the body, so the contact force has a backward component (1 true) and does negative work, dissipating energy (3 true).
2. The contact force must produce a torque about the centre to slow the rotation, so it does not pass through the centre (2 false).
3. Vertical equilibrium gives \( N=mg \) (4 false).

Teacher's Note:
1. A body that is slowing down needs a backward force and a retarding torque, so the contact force is offset from the centre.

Answer: (c) \( \dfrac{2}{3}ma^2 \)

Explanation:
1. About the centre: \( I_c=\dfrac{ma^2}{12}+\dfrac{ma^2}{12}=\dfrac{ma^2}{6} \).
2. The corner is at distance \( \dfrac{a}{\sqrt2} \) from the centre, so \( I=\dfrac{ma^2}{6}+m\dfrac{a^2}{2}=\dfrac23ma^2 \).

Teacher's Note:
1. Use the perpendicular axis theorem for the centre and then the parallel axis theorem to the corner.

Answer: (d) first increases and then decreases

Explanation:
1. Angular momentum is conserved: \( I\omega=\text{const} \).
2. The insect's distance from the axis first decreases to zero (at the centre) and then increases, so the total moment of inertia first decreases then increases.
3. Hence the angular speed first increases and then decreases.

Teacher's Note:
1. Watch how the insect's distance from the axis changes along the diameter.

Answer: (b) more than 3 but less than 6

Explanation:
1. \( \alpha=\dfrac{FR}{I}=\dfrac{2(20t-5t^2)}{10}=4t-t^2 \).
2. \( \omega=2t^2-\dfrac{t^3}{3} \), which is zero at \( t=6 \) s.
3. \( \theta=\int_0^6\omega\,dt=\left[\dfrac{2t^3}{3}-\dfrac{t^4}{12}\right]_0^6=144-108=36 \) rad, i.e. \( \dfrac{36}{2\pi}\approx5.7 \) rotations.

Teacher's Note:
1. Integrate the angular acceleration twice up to the time the angular speed becomes zero.

Answer: (b) \( \dfrac{r\omega_0}{2} \)

Explanation:
1. Angular momentum about a point on the ground is conserved: \( mr^2\omega_0=mvr+mr^2\omega \).
2. Rolling gives \( \omega=v/r \): \( mr^2\omega_0=2mvr \).
3. So \( v=\dfrac{r\omega_0}{2} \).

Teacher's Note:
1. Take moments about the ground, where friction has no torque, and use the rolling condition at the end.

Answer: (b) changes in direction but not in magnitude

Explanation:
1. The magnitude \( |\vec L| \) is fixed by the speed and geometry.
2. The direction of \( \vec L \) (slanted from the vertical) rotates with the bob, so \( \vec L \) is not constant.
3. So it changes in direction only.

Teacher's Note:
1. Gravity and tension give a net torque about the support that is horizontal and keeps turning the vector.